Question 1126865: Here is a multiplication involving two 16-digit numbers:
3851902343886132x5221791683705111
The 32-digit product is 20 113 831 625 748 8_8 690 240 050 420 652
What is the missing digit in the product?
Found 4 solutions by MathLover1, greenestamps, Alan3354, ikleyn:
Answer by MathLover1(20849)   (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
This could be a clever problem if there were some way to answer it without performing the entire calculation using divisibility rules.
For example, if each number were divisible by 3, then the product would be divisible by 9. Or if either number were divisible by 11, the product would be divisible by 11.
The rules of divisibility for 9 and 11 are simple enough that it would have been possible to find the answer with only a little work.
But I see no such shortcut, so it appears the answer must be found by performing the entire multiplication.
As such, it seems a useless problem, since it teaches nothing about problem solving....
BTW... the answer from the other tutor is correct.
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! Here is a multiplication involving two 16-digit numbers:
3851902343886132x5221791683705111
The 32-digit product is 20 113 831 625 748 8_8 690 240 050 420 652
What is the missing digit in the product?
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3851902343886132*5221791683705111
Using excess of 9's, or "casting out 9's" - I forget what it's called:
--> 3*5 = 15 --> 6
20 113 831 625 748 8_8 690 240 050 420 652 --> 4
The missing digit is 2.
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
To solve the problem, use the divisibility by 9 rule in this extended form:
The remainder of division the number by 9 is the same as the remainder of the sum of its digits divided by 9.
It is very widely known rule. See, for example, the lesson Divisibility by 9 rule in this site.
The number 3851902343886132 (first 16-digit factor) has the sum of digits 66.
Hence, according to the divisibility rule above, its remainder of division by 9 is the same
as the remainder of the number 66 divided by 9, i.e. 66-7*9 = 66-63 = 3.
The number 5221791683705111 (second 16-digit factor) has the sum of digits 59.
Hence, according to the divisibility rule above, its remainder of division by 9 is the same
as the remainder of the number 59 divided by 9, i.e. 59-6*9 = 59-54 = 5.
It implies that the remainder of the product of these two 16-digit numbers when divided by 9
is 3*5 (mod 9) = 15 (mod 9), i.e. 6.
The sum of digits of the 32-digit product 20 113 831 625 748 808 690 240 050 420 652
( ! I temporarily inserted "0" (zero) at the missed position for a moment)
is 112, which gives the remainder 4 when divided by 9.
We need to get the remainder 6, so I need to replace the inserted "0" by "2" to get it.
Hence, the missed digit under the question is "2".
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